Thermodynamics: An Engineering Approach ( 9th International Edition ) ISBN:9781260092684
Thermodynamics: An Engineering Approach ( 9th International Edition ) ISBN:9781260092684
9th Edition
ISBN: 9781260048667
Author: Yunus A. Cengel Dr.; Michael A. Boles
Publisher: McGraw-Hill Education
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Chapter 14.7, Problem 137RP

(a)

To determine

The temperature and relative humidity of the air when it leaves the heating section.

(a)

Expert Solution
Check Mark

Answer to Problem 137RP

The temperature is 33.1°C and relative humidity of the air when it leaves the heating section is 18.5%.

Explanation of Solution

As the process is a steady flow and thus the mass flow rate of dry air remains constant during the entire process.

m˙a1=m˙a2=m˙a

Here, the mass flow rate of air at inlet is m˙a1, mass flow rate of dry air at exit is m˙a2 and mass flow rate of dry air is m˙a.

The amount of moisture in the air remains constant as it flows through the heating section as process involves no dehumidification or humidification.

ω1=ω2 (I)

Here, specific humidity at state 1 and 2 is ω1andω2 respectively.

Express initial pressure of vapor.

Pν1=ϕ1Pg1=ϕ1Psat@15°C (II)

Here, relative humidity at state 1 is ϕ1, initial pressure of water vapor is Pg1 and saturation pressure at temperature of 15°C is Psat@15°C.

Express initial humidity ratio.

ω1=0.622Pν1P1Pν1 (III)

Here, pressure at state 1 is P1.

Express initial enthalpy.

h1=cpT1+ω1hg1@15°C (IV)

Here, specific heat at constant pressure is cp temperature at state 1 is T1 and initial specific enthalpy saturated vapor at temperature of 15°C is hg1@15°C.

Express specific volume at state 1.

v1=RaT1Pa1=RaT1P1Pν1 (V)

Here, gas constant of air is Ra, partial pressure of air at state 1 is Pa1 and temperature at state 1 is T1.

Express pressure vapor at state 3.

Pν3=ϕ3Pg3=ϕ3Psat@25°C (VI)

Here, relative humidity at state 3 is ϕ3, vapor pressure at state 3 is Pg3 and saturation pressure at temperature of 25°C is Psat@25°C.

Express humidity ratio at state 3.

ω3=0.622Pν3P3Pν3 (VII)

Here, pressure at state 3 is P3.

Express enthalpy at state 3.

h3=cpT3+ω3hg3@25°C (VIII)

Here, specific heat at constant pressure is cp, temperature at state 3 is T3 and specific enthalpy saturated vapor at state 3 and temperature of 25°C is hg3@25°C.

Express enthalpy at state 2.

h2h3 (IX)

Express enthalpy at state 2 to obtain the temperature at heating section.

h2=cpT2+ω3hg2=cpT2+ω3(2500.9+1.82T2) (X)

Here, temperature at heating section is T2, specific heat at constant pressure is cp, temperature at state 3 is T3 and specific enthalpy saturated vapor at state 3 and temperature of 25°C is hg3@25°C.

Express pressure of water vapor at state 2.

Pg2=Psat@T2 (XI)

Here, saturation pressure at temperature at leaving section is Psat@T2.

Express relative humidity at heating section.

ϕ2=ω2P2(0.622+ω2)Pg2 (XII)

Here, pressure at state 2 is P2.

Conclusion:

Refer Table A-4, “saturated water-temperature table”, and write the saturation pressure and initial specific enthalpy saturated vapor at temperature of 15°C.

Psat@15°C=1.7057kPahg1@15°C=2528.3kJ/kg

Substitute 0.55 for ϕ1 and 1.7057kPa for Psat@15°C in Equation (II).

Pν1=(0.55)(1.7057kPa)=0.938kPa

Substitute 0.938kPa for Pν1 and 96kPa for P1 in Equation (III).

ω1=0.622(0.938kPa)96kPa0.938kPa=0.006138kgH2O/kgdryair

Substitute 0.006138kgH2O/kgdryair for ω1 in Equation (I).

ω2=0.006138kgH2O/kgdryair

Refer Table A-2, “ideal-gas specific heats of various common gases”, and write the properties of air.

cp=1.005kJ/kg°CRa=0.287kPam3/kgK

Substitute 1.005kJ/kg°C for cp, 15°C for T1, 0.006138kgH2O/kgdryair for ω1 and 2528.3kJ/kg for hg1@15°C in Equation (IV).

h1=(1.005kJ/kg°C)(15°C)+(0.00638)(2528.3kJ/kg)=30.59kJ/kgdryair

Substitute 0.287kPam3/kgK for Ra, 15°C for T1, 0.938kPa for Pν1 and 96kPa for P1 in Equation (V).

v1=(0.287kPam3/kgK)(15°C)96kPa0.938kPa=(0.287kPam3/kgK)(15+273)K95.06kPa=(0.287kPam3/kgK)(288K)95.06kPa=0.8695m3/kgdryair

Refer Table A-4, “saturated water-temperature table”, and write the saturation pressure and initial specific enthalpy saturated vapor at temperature of 25°C.

Psat@25°C=3.17kPahg3@25°C=2546.5kJ/kg

Substitute 0.45 for ϕ3 and 3.17kPa for Psat@25°C in Equation (VI).

Pν3=(0.45)(3.17kPa)=1.426kPa

Substitute 1.426kPa for Pν3 and 96kPa for P3 in Equation (VII).

ω3=0.622(1.426kPa)96kPa1.426kPa=0.009381kgH2O/kgdryair

Substitute 1.005kJ/kg°C for cp, 25°C for T3, 0.009381kgH2O/kgdryair for ω3 and 2546.5kJ/kg for hg3@25°C in Equation (VIII).

h3=(1.005kJ/kg°C)(25°C)+(0.00938)(2546.5kJ/kg)=49.01kJ/kgdryair

Substitute 49.01kJ/kgdryair for h3 in Equation (IX).

h2=49.01kJ/kgdryair

Substitute 49.01kJ/kgdryair for h2, 0.006138kgH2O/kgdryair for ω2 and 1.005kJ/kg°C for cp in Equation (X).

49.01kJ/kgdryair=(1.005kJ/kg°C)T2+0.006138(2500.9+1.82T2)T2=33.1°C

Hence, the temperature of the air when it leaves the heating section is 33.1°C.

Substitute 33.1°C for T2 in Equation (XI).

Pg2=Psat@33.1°C (XIII)

Refer Table A-4, “saturated water-temperature table”, and write the saturation temperature or exit temperature at temperature of 33.1°C using an interpolation method.

Write the formula of interpolation method of two variables.

y2=(x2x1)(y3y1)(x3x1)+y1 (XIV)

Here, the variables denote by x and y is temperature and exit or saturation temperature respectively.

Show the saturation pressure corresponding to temperature as in Table (1).

Exit temperature

Tsat(°C)

Saturation pressure

Psat(kPa)

30 (x1)4.2469 (y1)
33.1 (x2)(y2=?)
35 (x3)5.6291 (y3)

Substitute 30°C,33.1°Cand35°C for x1,x2andx3 respectively, 4.2469kPa for y1 and 5.6291kPa for y3 in Equation (XIV).

y2=(33.1°C30°C)(5.6291kPa4.2469kPa)(35°C30°C)+4.2469kPa=5.072kPa=Psat@33.1°C

Substitute 5.072kPa for Psat@33.1°C in Equation (XIII).

Pg2=5.072kPa

Substitute 0.006138kgH2O/kgdryair for ω2, 96kPa for P2 and 5.072kPa for Pg2 in Equation (XII).

ϕ2=(0.006138)(96kPa)(0.622+0.006138)5.072kPa=0.185=18.5%

Hence, the relative humidity of the air when it leaves the heating section is 18.5%.

(b)

To determine

The rate of heat transfer in the heating section.

(b)

Expert Solution
Check Mark

Answer to Problem 137RP

The rate of heat transfer in the heating section is 636kJ/min.

Explanation of Solution

Express the rate of heat transfer in the heating section.

Q˙in=m˙a(h2h1) (XV)

Here, mass flow rate of air is m˙a.

Express mass flow rate of air.

m˙a=ν˙1v1 (XVI)

Here, volume flow rate at inlet is ν˙1.

Conclusion:

Substitute 30m3/min for ν˙1 and 0.8695m3/kg for v1 in Equation (XVI).

m˙a=30m3/min0.8695m3/kg=34.5kg/min

Substitute 34.5kg/min for m˙a, 49.01kJ/kgdryair for h2 and 30.59kJ/kgdryair for h1 in Equation (XV).

Q˙in=(34.5kg/min)[(49.0130.59)kJ/kgdryair]=636kJ/min

Hence, the rate of heat transfer in the heating section is 636kJ/min.

(c)

To determine

The rate of water added to air in the evaporative cooler.

(c)

Expert Solution
Check Mark

Answer to Problem 137RP

The rate of water added to air in the evaporative cooler is 0.112kg/min.

Explanation of Solution

Express the rate of water added to air in the evaporative cooler.

m˙w,added=m˙a(ω3ω2) (XVII)

Conclusion:

Substitute 0.009381kgH2O/kgdryair for ω3, 0.006138kgH2O/kgdryair for ω2 and 34.5kg/min for m˙a in Equation (XVII).

m˙w,added=(34.5kg/min)(0.0093810.006138)=0.112kg/min

Hence, the rate of water added to air in the evaporative cooler is 0.112kg/min.

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Chapter 14 Solutions

Thermodynamics: An Engineering Approach ( 9th International Edition ) ISBN:9781260092684

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